# Longest subsequence such that difference between adjacents is one

Given an array of n size, the task is to find the longest subsequence such that difference between adjacents is one.

Examples:

Input :arr[] = {10, 9, 4, 5, 4, 8, 6}Output :3 As longest subsequences with difference 1 are, "10, 9, 8", "4, 5, 4" and "4, 5, 6"Input :arr[] = {1, 2, 3, 2, 3, 7, 2, 1}Output :7 As longest consecutive sequence is "1, 2, 3, 2, 3, 2, 1"

This problem is based upon the concept of Longest Increasing Subsequence Problem.

Let arr[0..n-1] be the input array and dp[i] be the length of the longest subsequence (with differences one) ending at index i such that arr[i] is the last element of the subsequence. Then, dp[i] can be recursively written as: dp[i] = 1 + max(dp[j]) where 0 < j < i and [arr[j] = arr[i] -1 or arr[j] = arr[i] + 1] dp[i] = 1, if no such j exists. To find the result for a given array, we need to return max(dp[i]) where 0 < i < n.

Following is a Dynamic Programming based implementation. It follows the recursive structure discussed above.

## C++

// C++ program to find the longest subsequence such // the difference between adjacent elements of the // subsequence is one. #include <bits/stdc++.h> using namespace std; // Function to find the length of longest subsequence int longestSubseqWithDiffOne(int arr[], int n) { // Initialize the dp[] array with 1 as a // single element will be of 1 length int dp[n]; for (int i = 0; i < n; i++) dp[i] = 1; // Start traversing the given array for (int i = 1; i < n; i++) { // Compare with all the previous elements for (int j = 0; j < i; j++) { // If the element is consecutive then // consider this subsequence and update // dp[i] if required. if ((arr[i] == arr[j] + 1) || (arr[i] == arr[j] - 1)) dp[i] = max(dp[i], dp[j] + 1); } } // Longest length will be the maximum value // of dp array. int result = 1; for (int i = 0; i < n; i++) if (result < dp[i]) result = dp[i]; return result; } // Driver code int main() { // Longest subsequence with one difference is // {1, 2, 3, 4, 3, 2} int arr[] = { 1, 2, 3, 4, 5, 3, 2 }; int n = sizeof(arr) / sizeof(arr[0]); cout << longestSubseqWithDiffOne(arr, n); return 0; }

## Java

// Java program to find the longest subsequence // such that the difference between adjacent // elements of the subsequence is one. import java.io.*; class GFG { // Function to find the length of longest // subsequence static int longestSubseqWithDiffOne(int arr[], int n) { // Initialize the dp[] array with 1 as a // single element will be of 1 length int dp[] = new int[n]; for (int i = 0; i < n; i++) dp[i] = 1; // Start traversing the given array for (int i = 1; i < n; i++) { // Compare with all the previous // elements for (int j = 0; j < i; j++) { // If the element is consecutive // then consider this subsequence // and update dp[i] if required. if ((arr[i] == arr[j] + 1) || (arr[i] == arr[j] - 1)) dp[i] = Math.max(dp[i], dp[j] + 1); } } // Longest length will be the maximum // value of dp array. int result = 1; for (int i = 0; i < n; i++) if (result < dp[i]) result = dp[i]; return result; } // Driver code public static void main(String[] args) { // Longest subsequence with one // difference is // {1, 2, 3, 4, 3, 2} int arr[] = { 1, 2, 3, 4, 5, 3, 2 }; int n = arr.length; System.out.println(longestSubseqWithDiffOne( arr, n)); } } // This code is contributed by Prerna Saini

## Python

# Function to find the length of longest subsequence def longestSubseqWithDiffOne(arr, n): # Initialize the dp[] array with 1 as a # single element will be of 1 length dp = [1 for i in range(n)] # Start traversing the given array for i in range(n): # Compare with all the previous elements for j in range(i): # If the element is consecutive then # consider this subsequence and update # dp[i] if required. if ((arr[i] == arr[j]+1) or (arr[i] == arr[j]-1)): dp[i] = max(dp[i], dp[j]+1) # Longest length will be the maximum value # of dp array. result = 1 for i in range(n): if (result < dp[i]): result = dp[i] return result # Driver code arr = [1, 2, 3, 4, 5, 3, 2] # Longest subsequence with one difference is # {1, 2, 3, 4, 3, 2} n = len(arr) print longestSubseqWithDiffOne(arr, n) # This code is contributed by Afzal Ansari

## C#

// C# program to find the longest subsequence // such that the difference between adjacent // elements of the subsequence is one. using System; class GFG { // Function to find the length of longest // subsequence static int longestSubseqWithDiffOne(int[] arr, int n) { // Initialize the dp[] array with 1 as a // single element will be of 1 length int[] dp = new int[n]; for (int i = 0; i < n; i++) dp[i] = 1; // Start traversing the given array for (int i = 1; i < n; i++) { // Compare with all the previous // elements for (int j = 0; j < i; j++) { // If the element is consecutive // then consider this subsequence // and update dp[i] if required. if ((arr[i] == arr[j] + 1) || (arr[i] == arr[j] - 1)) dp[i] = Math.Max(dp[i], dp[j] + 1); } } // Longest length will be the maximum // value of dp array. int result = 1; for (int i = 0; i < n; i++) if (result < dp[i]) result = dp[i]; return result; } // Driver code public static void Main() { // Longest subsequence with one // difference is // {1, 2, 3, 4, 3, 2} int[] arr = { 1, 2, 3, 4, 5, 3, 2 }; int n = arr.Length; Console.Write( longestSubseqWithDiffOne(arr, n)); } } // This code is contributed by nitin mittal.

## PHP

<?php // PHP program to find the longest // subsequence such the difference // between adjacent elements of the // subsequence is one. // Function to find the length of // longest subsequence function longestSubseqWithDiffOne($arr, $n) { // Initialize the dp[] // array with 1 as a // single element will // be of 1 length $dp[$n] = 0; for($i = 0; $i< $n; $i++) $dp[$i] = 1; // Start traversing the // given array for($i = 1; $i < $n; $i++) { // Compare with all the // previous elements for($j = 0; $j < $i; $j++) { // If the element is // consecutive then // consider this // subsequence and // update dp[i] if // required. if (($arr[$i] == $arr[$j] + 1) || ($arr[$i] == $arr[$j] - 1)) $dp[$i] = max($dp[$i], $dp[$j] + 1); } } // Longest length will be // the maximum value // of dp array. $result = 1; for($i = 0 ; $i < $n ; $i++) if ($result < $dp[$i]) $result = $dp[$i]; return $result; } // Driver code // Longest subsequence with // one difference is // {1, 2, 3, 4, 3, 2} $arr = array(1, 2, 3, 4, 5, 3, 2); $n = sizeof($arr); echo longestSubseqWithDiffOne($arr, $n); // This code is contributed by nitin mittal. ?>

## Javascript

<script> // JavaScript program to find the // longest subsequence such that the // difference between adjacent elements // of the subsequence is one. // Function to find the length of longest // subsequence function longestSubseqWithDiffOne(arr, n) { // Initialize the dp[] array with 1 as a // single element will be of 1 length let dp = []; for(let i = 0; i < n; i++) dp[i] = 1; // Start traversing the given array for(let i = 1; i < n; i++) { // Compare with all the previous // elements for(let j = 0; j < i; j++) { // If the element is consecutive // then consider this subsequence // and update dp[i] if required. if ((arr[i] == arr[j] + 1) || (arr[i] == arr[j] - 1)) dp[i] = Math.max(dp[i], dp[j] + 1); } } // Longest length will be the maximum // value of dp array. let result = 1; for(let i = 0; i < n; i++) if (result < dp[i]) result = dp[i]; return result; } // Driver Code // Longest subsequence with one // difference is // {1, 2, 3, 4, 3, 2} let arr = [1, 2, 3, 4, 5, 3, 2]; let n = arr.length; document.write(longestSubseqWithDiffOne(arr, n)); // This code is contributed by souravghosh0416 </script>

**Output: **

6

**Time Complexity: **O(n^{2}) **Auxiliary Space: **O(n)

**Efficient Approach**

## C++

#include<bits/stdc++.h> using namespace std; int longestSubsequence(int n, int arr[]) { if(n==1) return 1; int dp[n]; unordered_map<int,int> mapp; dp[0]=1; mapp[arr[0]]=0; for(int i=1;i<n;i++){ if(abs(arr[i]-arr[i-1])==1) dp[i]=dp[i-1]+1; else{ if(mapp.count(arr[i]+1) >0 || mapp.count(arr[i]-1)){ dp[i]=1+max(mapp[arr[i]+1],mapp[arr[i]-1]); } else dp[i]=1; } mapp[arr[i]]=dp[i]; } return (*max_element(dp,dp+n)); } int main() { // Longest subsequence with one difference is // {1, 2, 3, 4, 3, 2} int arr[] = {1, 2, 3, 4, 5, 3, 2}; int n = sizeof(arr)/sizeof(arr[0]); cout << longestSubsequence(n, arr); return 0; }

## Java

import java.lang.Math; import java.util.*; class GFG { static int longestSubsequence(int n, int arr[]) { if (n == 1) return 1; Integer dp[] = new Integer[n]; HashMap<Integer, Integer> mapp = new HashMap<>(); dp[0] = 1; mapp.put(arr[0], 0); for (int i = 1; i < n; i++) { if (Math.abs(arr[i] - arr[i - 1]) == 1) dp[i] = dp[i - 1] + 1; else { if (mapp.containsKey(arr[i] + 1) || mapp.containsKey(arr[i] - 1)) { dp[i] = 1 + Math.max(mapp.getOrDefault( arr[i] + 1, 0), mapp.getOrDefault( arr[i] - 1, 0)); } else dp[i] = 1; } mapp.put(arr[i], dp[i]); } return Collections.max(Arrays.asList(dp)); } public static void main(String[] args) { // Longest subsequence with one // difference is // {1, 2, 3, 4, 3, 2} int arr[] = { 1, 2, 3, 4, 5, 3, 2 }; int n = arr.length; System.out.println(longestSubsequence(n, arr)); } } // This code is contributed by rajsanghavi9.

## Python3

def longestSubsequence(A, N): L = [1]*N hm = {} for i in range(1,N): if abs(A[i]-A[i-1]) == 1: L[i] = 1 + L[i-1] elif hm.get(A[i]+1,0) or hm.get(A[i]-1,0): L[i] = 1+max(hm.get(A[i]+1,0), hm.get(A[i]-1,0)) hm[A[i]] = L[i] return max(L) # Driver code A = [1, 2, 3, 4, 5, 3, 2] N = len(A) print(longestSubsequence(A, N))

**Output: **

6

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